The molarity of a 1 molal glucose solution h | vedclass

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(B) Given: Molality $(m)$ = $1 \ mol \ kg^{-1}$,Density $(d)$ = $1.2 \ g \ mL^{-1}$.Let the mass of the solvent (water) be $1000 \ g$ $(1 \ kg)$.The m

Vedclass · Accepted Answer

$1.01$

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(B) Given: Molality $(m)$ = $1 \ mol \ kg^{-1}$,Density $(d)$ = $1.2 \ g \ mL^{-1}$.Let the mass of the solvent (water) be $1000 \ g$ $(1 \ kg)$.The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ g \ mol^{-1}$.Since $m = 1 \ mol \ kg^{-1}$,the mass of glucose in $1 \ kg$ of water is $180 \ g$.Total mass of the solution = Mass of solute + Mass of solvent = $180 \ g + 1000 \ g = 1180 \ g$.Volume of the solution = $\frac{	ext{Mass of solution}}{	ext{Density}} = \frac{1180 \ g}{1.2 \ g \ mL^{-1}} = 983.33 \ mL = 0.9833 \ L$.Molarity $(M)$ = $\frac{	ext{Moles of solute}}{	ext{Volume of solution in } L} = \frac{1 \ mol}{0.9833 \ L} \approx 1.01 \ M$.

The molarity of a $1 \ molal$ glucose solution having a density of $1.2 \ g/mL$ is: (in $M$)

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